A Level Chemistry: Enthalpies
Sort of a random question but I understand that the bond enthalpy for each bond in a compound can be different even if its the same type of bond. So I wanted to ask if you had a molecule with 3 (e.g) X=Y bonds and you were breaking each of these bonds one after the other, would the final one take more energy to break?
Linking this to cyclic molecules: if you had cyclohexat-1,3-diene vs cyclohexa-1,4-diene, what would the approximate bond enthalpies be for the C=Cs in each case and why?
I hope my question makes sense
Linking this to cyclic molecules: if you had cyclohexat-1,3-diene vs cyclohexa-1,4-diene, what would the approximate bond enthalpies be for the C=Cs in each case and why?
I hope my question makes sense
A difficult question. Firstly, average bond enthalpies are calculated using a wide variety of molecules in the gas phase. They don't really give any useful information over which isomer is more stable than another. The two compounds you mentioned have exactly the same number and type of bonds, but in different places. Mean bond enthalpy data would give the same value for both.
We can do direct experiments on each compound and measure the enthalpy of hydrogenation. In this case, you will find that cyclohexa-1,3-diene is slightly more stable than cyclohexa-1,4-diene. The 1,3-diene has the two double bonds separated by a single C-C bond - this is called 'conjugation'. All conjugated dienes are stabilized through delocalization of the electrons. In the 1,4 diene the C=C are too far apart for this to occur.
We can do direct experiments on each compound and measure the enthalpy of hydrogenation. In this case, you will find that cyclohexa-1,3-diene is slightly more stable than cyclohexa-1,4-diene. The 1,3-diene has the two double bonds separated by a single C-C bond - this is called 'conjugation'. All conjugated dienes are stabilized through delocalization of the electrons. In the 1,4 diene the C=C are too far apart for this to occur.
(edited 1 year ago)
Original post by tony_dolby
A difficult question. Firstly, average bond enthalpies are calculated using a wide variety of molecules in the gas phase. They don't really give any useful information over which isomer is more stable than another. The two compounds you mentioned have exactly the same number and type of bonds, but in different places. Mean bond enthalpy data would give the same value for both.
We can do direct experiments on each compound and measure the enthalpy of hydrogenation. In this case, you will find that cyclohexa-1,3-diene is slightly more stable than cyclohexa-1,4-diene. The 1,3-diene has the two double bonds separated by a single C-C bond - this is called 'conjugation'. All conjugated dienes are stabilized through delocalization of the electrons. In the 1,4 diene the C=C are too far apart for this to occur.
We can do direct experiments on each compound and measure the enthalpy of hydrogenation. In this case, you will find that cyclohexa-1,3-diene is slightly more stable than cyclohexa-1,4-diene. The 1,3-diene has the two double bonds separated by a single C-C bond - this is called 'conjugation'. All conjugated dienes are stabilized through delocalization of the electrons. In the 1,4 diene the C=C are too far apart for this to occur.
Hi, thank you for your reply. You know how you said the mean bond enthalpies will be the same for both because its the same type of bond. What I wanted to know was how would the actual bond enthalpies compare? As in, in what way does its position affect its true bond enthalpy - when is it likely to be less/more than the mean?
And secondly, thank you for telling me about conjugation. does this relate to overlapping p orbitals?
Original post by fluffypoopies
Hi, thank you for your reply. You know how you said the mean bond enthalpies will be the same for both because its the same type of bond. What I wanted to know was how would the actual bond enthalpies compare? As in, in what way does its position affect its true bond enthalpy - when is it likely to be less/more than the mean?
And secondly, thank you for telling me about conjugation. does this relate to overlapping p orbitals?
And secondly, thank you for telling me about conjugation. does this relate to overlapping p orbitals?
Gosh, now you've got me. The easy answer is that there is some overlapping of the Pi system which increases stability.
The question of bond energy is more difficult. The general consensus is that shorter bonds are stronger. In 1,3 dienes the single bond is shorter and stronger (147 pm) than an ordinary alkane single bond (154 pm). This bond has significant double bond character. Common sense has always suggested that the C=C bonds in conjugated dienes would be longer to compensate. It seems that this is indeed the case.
https://cen.acs.org/articles/84/i23/Butadienes-Bonds.html
(edited 1 year ago)
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